EconomicsISC Class 11 · Correlation

Spearman's Rank Correlation and Choosing the Method (ISC Class 11 Economics)

Updated 17 October 2026 · Fact-checked

Spearman's rank correlation coefficient R = 1 - 6 Sum d squared / [n(n squared - 1)], where d is the difference between the ranks of each pair and n is the number of pairs. It is used when data are ranks or qualitative, or when the numerical data have outliers. For tied ranks, add m(m squared - 1)/12 to Sum d squared for each group of m tied values. R lies between -1 and +1.

Tied values share the average of the ranks they would take (2 and 3 give 2.5 each), then add m(m²−1)/12 to Σd² for each tied group.

Illustration drawn for QuestionBuddy.

Understand Spearman's Rank Correlation, Tied Ranks and Choosing the Method

When to use. When the variables cannot be measured exactly but can be ranked (beauty contest, ability), when only ranks are given, or when the data have extreme values. Pearson's r is better for accurately measured quantitative data.

Ranking. Give rank 1 to the highest (or the lowest) value in each variable, consistently for both. Find d = Rank X - Rank Y for each pair.

Formula (no ties).
R = 1 - [6 x Sum d squared] / [n x (n squared - 1)]
Check: Sum d = 0 always.

Tied (repeated) values. If two or more items have the same value, each gets the average of the ranks they would have occupied (two values tied for 2nd and 3rd each get 2.5). Then add a correction factor m(m squared - 1)/12 to Sum d squared for every group of m tied items (in both series):
R = 1 - 6[Sum d squared + Sum of m(m squared - 1)/12] / [n(n squared - 1)].

Interpretation. +1: perfect agreement of ranks; -1: perfect disagreement; 0: no relationship; same scale as r (above 0.75 high, 0.25 to 0.75 moderate, below 0.25 low).

Pearson vs Spearman.

Point	Pearson	Spearman
Data	Measured values	Ranks, or values converted to ranks
Outliers	Affected strongly	Little affected
Relationship	Linear	Monotonic order
Work	More arithmetic	Simpler for small n

Why Spearman can differ from Pearson. Converting values to ranks loses information about the size of differences.

Spearman's Rank Correlation, Tied Ranks and Choosing the Method: key facts

Spearman, no ties: R = 1 - 6 Sum d^2 / [n(n^2 - 1)] | d = Rank X - Rank Y; Sum d = 0.
Correction for ties: Correction = m(m^2 - 1)/12 | For each group of m tied items, added to Sum d^2; Applied in both series.
Tied rank: Rank of tied items = Average of the ranks they occupy | E.g. ranks 2 and 3 give 2.5; Rule.

How to answer Spearman's Rank Correlation, Tied Ranks and Choosing the Method questions

Rank both series in the same direction and tabulate d and d squared.

1Rank X and Y (1 for the highest).
2Compute d and d squared for each pair; check that Sum d = 0.
3For ties, assign average ranks and note the corrections.
4Substitute in the formula and interpret.

Spearman's Rank Correlation, Tied Ranks and Choosing the Method in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Judges' agreement

Suppose two judges rank five contestants as: Judge 1: 1, 2, 3, 4, 5 and Judge 2: 2, 1, 4, 3, 5.

How strongly do the judges agree?

Try it first, then show the solutionSolution

d = -1, 1, -1, 1, 0; d squared = 1, 1, 1, 1, 0; Sum = 4.
R = 1 - 6 x 4 / (5 x 24) = 1 - 24/120 = 0.8.

Answer: R = 0.8: high agreement.

Memory hook: Small d, high R.

Tricks to remember Spearman's Rank Correlation, Tied Ranks and Choosing the Method

Short memory aids. Say each one aloud once, then use it on a question the same day.

Six over n(n squared minus one)
R = 1 - 6 Sum d squared / n(n squared - 1).
Use it when: Formula.
Average the tied ranks
Tie for 2nd and 3rd: both 2.5.
Use it when: Ties.
Sum of d is zero
Use as a check.
Use it when: Accuracy.

Spearman's Rank Correlation, Tied Ranks and Choosing the Method: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Perfect agreement · 1 mark · MCQBoard level

If the two sets of ranks are identical, the rank correlation coefficient is:
(a) 0 (b) -1 (c) +1 (d) 0.5

Try it first, then show the solutionSolution

d = 0 for all pairs, so R = 1.

Answer: (c) +1

Example 2: Compute R · 6 marks · Long answerHarder board level

Six students' marks in two subjects: X = 45, 60, 55, 70, 50, 65 and Y = 40, 62, 58, 72, 48, 60. Find the rank correlation coefficient. (6 marks)

Try it first, then show the solutionSolution

Rank X (highest = 1): 45 → 6, 60 → 3, 55 → 4, 70 → 1, 50 → 5, 65 → 2.
Rank Y: 40 → 6, 62 → 2, 58 → 4, 72 → 1, 48 → 5, 60 → 3.
d = 0, 1, 0, 0, 0, -1; d squared = 0, 1, 0, 0, 0, 1; Sum d squared = 2.
R = 1 - 6 x 2 / (6 x 35) = 1 - 12/210 = 1 - 0.057 = 0.943.

Answer: R = 0.943: very high positive correlation.

Example 3: Tied ranks · 6 marks · Long answerHarder board level

Find R for X = 10, 20, 20, 30 and Y = 5, 15, 25, 25 (ranks with highest = 1; use the tie correction). (6 marks)

Try it first, then show the solutionSolution

Rank X: 30 → 1; the two 20s share ranks 2 and 3, each 2.5; 10 → 4. Ranks of the pairs: 4, 2.5, 2.5, 1.
Rank Y: the two 25s share ranks 1 and 2, each 1.5; 15 → 3; 5 → 4. Ranks: 4, 3, 1.5, 1.5.
d = 0, -0.5, 1, -0.5; d squared = 0, 0.25, 1, 0.25; Sum = 1.5.
Corrections: X has one tie of 2: 2 x 3/12 = 0.5; Y one tie of 2: 0.5. Total = 1.5 + 0.5 + 0.5 = 2.5.
R = 1 - 6 x 2.5 / (4 x 15) = 1 - 15/60 = 0.75.

Answer: R = 0.75: high positive correlation.

Example 4: Missing d squared · 4 marks · Short answerHarder board level

For 8 pairs, Sum d squared is 24. Find R. If one pair's d is mis-recorded as 4 but is actually 2, find the corrected R. (4 marks)

Try it first, then show the solutionSolution

R = 1 - 6 x 24 / (8 x 63) = 1 - 144/504 = 0.714.
Correct d squared: 16 replaced by 4, so Sum = 24 - 16 + 4 = 12.
R = 1 - 6 x 12 / 504 = 1 - 72/504 = 0.857.

Answer: R = 0.714 initially; corrected R = 0.857.

Example 5: Choose the method · 4 marks · Short answerHOTS / competency

A researcher has: (i) marks of 6 students in two tests; (ii) 8 candidates ranked by two interviewers; (iii) heights and weights of 40 persons, one of whom is a very tall outlier. Choose Pearson or Spearman for each with reasons. (4 marks)

Try it first, then show the solutionSolution

(i) Pearson is possible since marks are measured; Spearman is also acceptable for a small sample.
(ii) Spearman: only ranks are available.
(iii) Spearman reduces the effect of the outlier; Pearson would be pulled by the extreme value.

Answer: (i) Pearson (or Spearman); (ii) Spearman; (iii) Spearman for the outlier.

Where students lose marks in Spearman's Rank Correlation, Tied Ranks and Choosing the Method

The slips examiners see most often, and what to write instead.

Mistake: Ranking one series in each direction
Inconsistent ranking.
Wrong working
X highest = 1, Y lowest = 1
Right working
Both highest = 1
Do this: Use the same rule.
Usually costs: 2 marks.
Mistake: Forgetting tie corrections
Skipping a step.
Wrong working
R = 1 - 6 x 1.5 / 60
Right working
Add 0.5 for each tied group
Do this: List the ties first.
Usually costs: 2 marks.
Mistake: Not checking Sum d = 0
Rushing.
Wrong working
Sum d = 2 unnoticed
Right working
Sum d must be 0
Do this: Check before squaring.
Usually costs: 1-2 marks.

Exam tips

Write the ranks in a table with d and d squared.
Show the tie correction in a separate line.
Interpret R in a sentence.

Practice questions on Spearman's Rank Correlation, Tied Ranks and Choosing the Method

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Spearman's coefficient is mainly used when the data are:
(a) in the form of ranks (b) always very large (c) in millions (d) only negative
Hint
Order, not size.
Answer
(a) in the form of ranks
Q2 · 2 marks · Short answerBoard level
Two tied values would occupy the 3rd and 4th ranks. What rank is given to each, and what correction is added to Sum d squared? (2 marks)
Hint
Average and m(m squared - 1)/12.
Answer
Each gets (3 + 4) / 2 = 3.5; correction = 2 x (4 - 1) / 12 = 0.5.
Q3 · 4 marks · Short answerHarder board level
Five candidates are ranked by two examiners: Examiner A: 1, 2, 3, 4, 5; Examiner B: 3, 1, 2, 5, 4. Find the rank correlation coefficient. (4 marks)
Hint
d = -2, 1, 1, -1, 1.
Answer
d = -2, 1, 1, -1, 1; d squared = 4, 1, 1, 1, 1; Sum = 8. R = 1 - 6 x 8 / (5 x 24) = 1 - 48/120 = 0.6.
Q4 · 7 marks · Long answerHarder board level
Marks of 7 students in economics and statistics are: X = 82, 75, 75, 60, 91, 68, 55 and Y = 78, 80, 70, 62, 88, 70, 58. Find R with the tie correction (highest = rank 1). (7 marks)
Hint
Ties in both series.
Answer
Rank X: 91 → 1, 82 → 2, 75 and 75 → 3.5 each, 68 → 5, 60 → 6, 55 → 7. Rank Y: 88 → 1, 80 → 2, 78 → 3, 70 and 70 → 4.5 each, 62 → 6, 58 → 7. Pairs (X rank, Y rank): (2, 3), (3.5, 2), (3.5, 4.5), (6, 6), (1, 1), (5, 4.5), (7, 7). d = -1, 1.5, -1, 0, 0, 0.5, 0; d squared = 1, 2.25, 1, 0, 0, 0.25, 0; Sum = 4.5. Corrections: X one tie of 2: 0.5; Y one tie of 2: 0.5. Total = 5.5. R = 1 - 6 x 5.5 / (7 x 48) = 1 - 33/336 = 0.902.
Q5 · 3 marks · Short answerHOTS / competency
Pearson's r for a data set is 0.62 but Spearman's R is 0.93. One observation is a very large outlier. Explain the difference and which figure better describes the general relation. (3 marks)
Hint
Outlier effects.
Answer
The outlier distorts Pearson's r because it uses the size of deviations, while ranks reduce the outlier to a single largest rank. The ranks of the other items agree closely, so Spearman's 0.93 better describes the general relation.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Spearman formula?Show answer
R = 1 - 6 Sum d squared / n(n squared - 1).
2.Tied ranks get?Show answer
The average rank.
3.Tie correction?Show answer
m(m squared - 1)/12.
4.Sum of d equals?Show answer
Zero.
5.Use Spearman when?Show answer
Data are ranks or have outliers.

Spearman's Rank Correlation, Tied Ranks and Choosing the Method: frequently asked questions

Can R be -1?

Yes, for perfectly opposite ranks.

Rank highest as 1 or lowest as 1?

Either, if consistent.

Is R the same as r?

Not usually, same range though.

What if n is large?

Pearson may be easier by software.